Vector geometry: Proof of a trapezium and cross and dot product help

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Homework Help Overview

The discussion revolves around vector geometry, specifically focusing on cross and dot products, as well as the properties of trapeziums defined by vectors. Participants are exploring calculations involving vector magnitudes and relationships, as well as the implications of vector operations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to clarify the definitions and properties of the cross product and dot product, questioning how to interpret the squaring of a vector. There is also discussion about the conditions for a shape to be classified as a trapezium based on vector definitions.

Discussion Status

The conversation is ongoing, with participants providing insights and corrections regarding vector operations. Some have offered guidance on interpreting vector products, while others are still grappling with the concepts and seeking further clarification.

Contextual Notes

Participants express confusion regarding vector operations and their implications, particularly in relation to homework constraints and the definitions of geometric properties. There is a noted lack of consensus on how to approach the problems presented.

FilipaE
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Homework Statement


#1 Given that the angle between the vectors a and b is 2Pi/3 and |a|=3 and |b|=4 calculate:
(axb)^2 [(2a+b)x(a+2b)]^2

#2 Given three unit vectors, a, b, c such that a+b+c=0 find (a dot b) + (b dot c) + (c dot a)

#3 Given AB=a+2b BC=-4a-b CD= -5a-3b where a and b are any two vectors, prove that ABCD is a trapezium


Homework Equations



Given above

The Attempt at a Solution


#1 I know that axb = |a||b|sinx
and axa = 0 and axb = -b x a but i do not understand how the powers or constants affect this?

#2 No idea

#3 A trapezium has one set of parallel sides only and i know if vectors are parallel they are a multiple of each other, however none of these seem to be multiples of each other?
 
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FilipaE said:
#1 I know that axb = |a||b|sinx

This is false. The length of a\times b equals |a||b|\sin\theta.

Anyway, what do you mean with (a\times b)^2?? What does it mean to take the square of a vector??
 
I am using x as the cross product, meaning the cross product of a and b is |a||b|sinx, is this not correct?
 
and (a x b)^2 means the cross product of a and b all sqaured
 
FilipaE said:
I am using x as the cross product, meaning the cross product of a and b is |a||b|sinx, is this not correct?

No, that is not correct. Review your definition of cross product.
 
FilipaE said:
and (a x b)^2 means the cross product of a and b all sqaured

Yes, I get that: it means that you take the square of the cross product. But the cross product is a vector. What do you mean with the square of a vector?
 
This is the definition of cross product online |a x b|=|a||b|sinx
I have no idea what you mean vectors is a very confusing topic for me
 
FilipaE said:
This is the definition of cross product online |a x b|=|a||b|sinx
I have no idea what you mean vectors is a very confusing topic for me

Yes, the definition |a\times b|=|a||b|\sin x is correct. But writing a\times b=|a||b|sin x is not correct. Do you see the difference?
 
Oh i see, so how does the formula differ without the modulus sign? and what happens where there is addition involved as in Q1 part 2
 
  • #10
Do you understand that a\times b is a vector and that |a\times b| is a scalar??
 
  • #11
yes i do now, but given this information how would i write the vector a x b?
 
  • #12
Before you do that, you got to understand what (a\times b)^2 is. What does it mean to square a vector?
 
  • #13
Is it the dot product of the two vectors, creating a scalar?
 
  • #14
FilipaE said:
Is it the dot product of the two vectors, creating a scalar?

Yes.

So you need to find (a\times b)\cdot (a\times b).

Given a vector v, what is the definition of v.v?
 
  • #15
|v|^2?
 
  • #16
So (a\times b)\cdot (a\times b)=|a\times b|^2. Now you should be able to calculate this.
 
  • #17
Thankyou! I have got the answer 108?

So for [(2a+b) x(a+2b)]^2 I do [(2a+b)x(a+2b)] dot [(2a+b)x(a+2b)] = |(2a+b)x(a+2b)|^2

I know to do this i need to use the distributive law. Is it

2a x a + 2a x2b + b xa + b x2b?
Where 2a x a = 0 and bx2b=0?
 
  • #18
So the answer is 108? Thankyou!
For [(2a+b) x(a+2b)]^2 is this the same as
(2a+b) x(a+2b) dot (2a+b) x(a+2b) = |(2a+b) x(a+2b)|^2 ??

I know i have to use some sort of distributive law here but i am not sure on how to do it would it be 2a x a + 2a x 2b + b x a + b x 2b ? where 2a x a = 0 and b x 2b =0??
 
  • #19
Sorry didnt think the first message had sent!
 

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